First off, I apologize if this is in the wrong section of StackExchange, it's at the intersection of several different topics (mathematics, physics, and computer science), so I chose the section that seemed most relevant.
In my high school AP physics (not calculus-based) class, we are learning about electrostatics (electric fields and electromagnetic forces). For fun, I have decided to write a computer program for visualizing equipotential lines (these are lines such that all the points on the line have the same net electric potential) for two-dimensional situations.
The (simple) formula for the electric potential for a situation where there is only one charged object is:
$$electric \ potential = \frac{k*Q_1}{r_1}$$
where $k$ is a constant, $Q_1$ is the charge of the charged object, and $r_1$ is the distance from the charged object.
For use in a Cartesian plane, I changed the formula slightly, to
$$electrical \ potential = \frac{k*Q_1}{\sqrt{(x_{c1} – x)^2 + (y_{c1} – y)^2}}$$
where ($x_{c1}$, $y_{c1}$) is the location of the charged object (we're assuming the object is a point here)
Thus, an equipotential line when there are $n$ charged objects is the set of all points (x,y) such that
$$C = \sum_{i = 1} ^n \frac{k*Q_i}{\sqrt{(x_{c i} – x)^2 + (y_{c i} – y)^2}} $$
where $C$ is the electric potential at every point on the equipotential line.
Here's an example of several equipotential lines for a situation where there are 3 charged objects (plotted in Grapher)
I have tried simplifying the formula for the case where there are 3 charged objects (into something I know how to find solutions for) to no avail.
How can I solve this type of formula by hand? I'm most interested in cases where n = 3 (there are three charged objects), but if there's a more general solution to the problem, that would be awesome too.
I'm planning on implementing this in a computer program eventually, so having to do guess-and-check is fine. Details of the computer implementation are a question for later though, I'm mostly interested in understanding how to solve these types of relations.
EDIT:
Let me restate the question in a better way: I am looking to solve the formula for (x,y) pairs given $C$, and the $Q_i$ and ($x_{c i}$, $y_{c i}$) for each point.
Best Answer
Short answer: You've solved it as best as you can. Just add up the potentials as you've done.
Longer answer: If you take three (or more) arbitrary points, there's no symmetry to be exploited that results in a simpler equation. One point has radial symmetry, as the potential depends only on the distance from the charge. Two points have axial symmetry (there's an axis defined by the line joining the two points). Three points? No such luck. There may be some if you have an equilateral triangle or an isosceles triangle, but that's not the general case.